Merge pull request #694 from LetMeFly666/3019

添加问题“3019.按键变更的次数”的代码和题解

Author: LetMeFly666

Codes/3019-number-of-changing-keys.cpp

@@ -0,0 +1,20 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-01-07 13:03:56
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-01-07 13:08:37
+ */
+#ifdef _WIN32
+#include "_[1,2]toVector.h"
+#endif
+
+class Solution {
+public:
+    int countKeyChanges(string& s) {
+        int ans = 0;
+        for (int i = 1; i < s.size(); i++) {
+            ans += tolower(s[i]) != tolower(s[i - 1]);
+        }
+        return ans;
+    }
+};

Codes/3019-number-of-changing-keys.go

@@ -0,0 +1,17 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-01-07 13:11:57
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-01-07 13:13:25
+ */
+package main
+import "strings"
+
+func countKeyChanges(s string) (ans int) {
+    for i := 1; i < len(s); i++ {
+        if strings.ToLower(string(s[i])) != strings.ToLower(string(s[i - 1])) {
+            ans++
+        }
+    }
+    return
+}

Codes/3019-number-of-changing-keys.java

@@ -0,0 +1,17 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-01-07 13:10:05
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-01-07 13:11:25
+ */
+class Solution {
+    public int countKeyChanges(String s) {
+        int ans = 0;
+        for (int i = 1; i < s.length(); i++) {
+            if (Character.toLowerCase(s.charAt(i)) != Character.toLowerCase(s.charAt(i - 1))) {
+                ans++;
+            }
+        }
+        return ans;
+    }
+}

Codes/3019-number-of-changing-keys.py

@@ -0,0 +1,9 @@
+'''
+Author: LetMeFly
+Date: 2025-01-07 13:09:14
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-01-07 13:09:43
+'''
+class Solution:
+    def countKeyChanges(self, s: str) -> int:
+        return sum(s[i].lower() != s[i - 1].lower() for i in range(1, len(s)))

README.md

@@ -696,6 +696,7 @@
 |2974.最小数字游戏|简单|<a href="https://leetcode.cn/problems/minimum-number-game/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/07/12/LeetCode%202974.%E6%9C%80%E5%B0%8F%E6%95%B0%E5%AD%97%E6%B8%B8%E6%88%8F/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/140365205" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/minimum-number-game/solutions/2840437/letmefly-2974zui-xiao-shu-zi-you-xi-pai-ib5em/" target="_blank">LeetCode题解</a>|
 |2982.找出出现至少三次的最长特殊子字符串II|中等|<a href="https://leetcode.cn/problems/find-longest-special-substring-that-occurs-thrice-ii/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/05/30/LeetCode%202982.%E6%89%BE%E5%87%BA%E5%87%BA%E7%8E%B0%E8%87%B3%E5%B0%91%E4%B8%89%E6%AC%A1%E7%9A%84%E6%9C%80%E9%95%BF%E7%89%B9%E6%AE%8A%E5%AD%90%E5%AD%97%E7%AC%A6%E4%B8%B2II/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/139334864" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-longest-special-substring-that-occurs-thrice-ii/solutions/2795996/letmefly-2982zhao-chu-chu-xian-zhi-shao-9ubg1/" target="_blank">LeetCode题解</a>|
 |3011.判断一个数组是否可以变为有序|中等|<a href="https://leetcode.cn/problems/find-if-array-can-be-sorted/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/07/13/LeetCode%203011.%E5%88%A4%E6%96%AD%E4%B8%80%E4%B8%AA%E6%95%B0%E7%BB%84%E6%98%AF%E5%90%A6%E5%8F%AF%E4%BB%A5%E5%8F%98%E4%B8%BA%E6%9C%89%E5%BA%8F/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/140391465" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-if-array-can-be-sorted/solutions/2841665/letmefly-3011pan-duan-yi-ge-shu-zu-shi-f-i9ck/" target="_blank">LeetCode题解</a>|
+|3019.按键变更的次数|简单|<a href="https://leetcode.cn/problems/number-of-changing-keys/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/01/07/LeetCode%203019.%E6%8C%89%E9%94%AE%E5%8F%98%E6%9B%B4%E7%9A%84%E6%AC%A1%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/144983704" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/number-of-changing-keys/solutions/3040949/letmefly-3019an-jian-bian-geng-de-ci-shu-pgzx/" target="_blank">LeetCode题解</a>|
 |3033.修改矩阵|简单|<a href="https://leetcode.cn/problems/modify-the-matrix/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/07/05/LeetCode%203033.%E4%BF%AE%E6%94%B9%E7%9F%A9%E9%98%B5/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/140219034" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/modify-the-matrix/solutions/2832430/letmefly-3033xiu-gai-ju-zhen-yuan-di-xiu-e0cy/" target="_blank">LeetCode题解</a>|
 |3038.相同分数的最大操作数目I|简单|<a href="https://leetcode.cn/problems/maximum-number-of-operations-with-the-same-score-i/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/06/07/LeetCode%203038.%E7%9B%B8%E5%90%8C%E5%88%86%E6%95%B0%E7%9A%84%E6%9C%80%E5%A4%A7%E6%93%8D%E4%BD%9C%E6%95%B0%E7%9B%AEI/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/139535702" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/maximum-number-of-operations-with-the-same-score-i/solutions/2804245/letmefly-3038xiang-tong-fen-shu-de-zui-d-dkj7/" target="_blank">LeetCode题解</a>|
 |3046.分割数组|简单|<a href="https://leetcode.cn/problems/split-the-array/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/12/28/LeetCode%203046.%E5%88%86%E5%89%B2%E6%95%B0%E7%BB%84/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/144789679" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/split-the-array/solutions/3032725/letmefly-3046fen-ge-shu-zu-ji-shu-by-tis-33vv/" target="_blank">LeetCode题解</a>|

Solutions/LeetCode 3019.按键变更的次数.md

@@ -0,0 +1,141 @@
+---
+title: 3019.按键变更的次数
+date: 2025-01-07 13:14:51
+tags: [题解, LeetCode, 简单, 字符串]
+---
+
+# 【LetMeFly】3019.按键变更的次数：遍历（转小写）
+
+力扣题目链接：[https://leetcode.cn/problems/number-of-changing-keys/](https://leetcode.cn/problems/number-of-changing-keys/)
+
+<p>给你一个下标从<strong> 0</strong> 开始的字符串 <code>s</code> ，该字符串由用户输入。按键变更的定义是：使用与上次使用的按键不同的键。例如 <code>s = "ab"</code> 表示按键变更一次，而 <code>s = "bBBb"</code> 不存在按键变更。</p>
+
+<p>返回用户输入过程中按键变更的次数。</p>
+
+<p><strong>注意：</strong><code>shift</code> 或 <code>caps lock</code> 等修饰键不计入按键变更，也就是说，如果用户先输入字母 <code>'a'</code> 然后输入字母 <code>'A'</code> ，不算作按键变更。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "aAbBcC"
+<strong>输出：</strong>2
+<strong>解释：</strong> 
+从 s[0] = 'a' 到 s[1] = 'A'，不存在按键变更，因为不计入 caps lock 或 shift 。
+从 s[1] = 'A' 到 s[2] = 'b'，按键变更。
+从 s[2] = 'b' 到 s[3] = 'B'，不存在按键变更，因为不计入 caps lock 或 shift 。
+从 s[3] = 'B' 到 s[4] = 'c'，按键变更。
+从 s[4] = 'c' 到 s[5] = 'C'，不存在按键变更，因为不计入 caps lock 或 shift 。
+</pre>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "AaAaAaaA"
+<strong>输出：</strong>0
+<strong>解释：</strong> 不存在按键变更，因为这个过程中只按下字母 'a' 和 'A' ，不需要进行按键变更。<!-- notionvc: 8849fe75-f31e-41dc-a2e0-b7d33d8427d2 -->
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 100</code></li>
+	<li><code>s</code> 仅由英文大写字母和小写字母组成。</li>
+</ul>
+
+
+    
+## 解题方法：遍历
+
+从第二个字符开始遍历字符串，如果当前字符串的小写和前一个字符的小写不相同，则答案数量加一。
+
++ 时间复杂度$O(len(s))$
++ 空间复杂度$O(1)$
+
+### AC代码
+
+#### C++
+
+```cpp
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-01-07 13:03:56
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-01-07 13:08:37
+ */
+class Solution {
+public:
+    int countKeyChanges(string& s) {
+        int ans = 0;
+        for (int i = 1; i < s.size(); i++) {
+            ans += tolower(s[i]) != tolower(s[i - 1]);
+        }
+        return ans;
+    }
+};
+```
+
+#### Python
+
+```python
+'''
+Author: LetMeFly
+Date: 2025-01-07 13:09:14
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-01-07 13:09:43
+'''
+class Solution:
+    def countKeyChanges(self, s: str) -> int:
+        return sum(s[i].lower() != s[i - 1].lower() for i in range(1, len(s)))
+```
+
+#### Java
+
+```java
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-01-07 13:10:05
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-01-07 13:11:25
+ */
+class Solution {
+    public int countKeyChanges(String s) {
+        int ans = 0;
+        for (int i = 1; i < s.length(); i++) {
+            if (Character.toLowerCase(s.charAt(i)) != Character.toLowerCase(s.charAt(i - 1))) {
+                ans++;
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### Go
+
+```go
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-01-07 13:11:57
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-01-07 13:13:25
+ */
+package main
+import "strings"
+
+func countKeyChanges(s string) (ans int) {
+    for i := 1; i < len(s); i++ {
+        if strings.ToLower(string(s[i])) != strings.ToLower(string(s[i - 1])) {
+            ans++
+        }
+    }
+    return
+}
+```
+
+> 同步发文于CSDN和我的[个人博客](https://blog.letmefly.xyz/)，原创不易，转载经作者同意后请附上[原文链接](https://blog.letmefly.xyz/2025/01/07/LeetCode%203019.%E6%8C%89%E9%94%AE%E5%8F%98%E6%9B%B4%E7%9A%84%E6%AC%A1%E6%95%B0/)哦~
+>
+> Tisfy：[https://letmefly.blog.csdn.net/article/details/144983704](https://letmefly.blog.csdn.net/article/details/144983704)

Solutions/Other-Accumulation-Messy.md

@@ -67,6 +67,19 @@ q.pop();
 
 有个**投机取巧但不值得做**的办法：删除“当前设备”并重新备份，系统会首先备份其他数据（正好是你想要的），最后备份“应用列表”。备份应用列表的过程中空间满了，备份终止，但你想备份的数据也已经备份完了。
 
+## 安卓应用独占蓝牙信道
+
+使用蓝牙耳机进行腾讯系列软件通话时（包括但不限于腾讯会议、微信视频），软件会独占蓝牙信道，使得：
+
+1. 蓝牙无法同时播放音乐
+2. 蓝牙声音奇大无比震耳欲聋，调到最小也很大，想调到0调不到
+
+电脑上腾讯系软件也这样，但是可用通过`在声音图标上右键 -> 声音 -> 播放 -> 双击蓝牙设备 -> 高级 -> 取消勾选“允许应用程序独占控制该设备”`解决。
+
+安卓上暂未找到可行办法。。。emm，我说你个tx，不给用户一个选项么，上来就独占，如果我不想让你独占呢。。。
+
+还有使用腾讯会议时经常弹出的“监测到音量过小，建议调整到50%以上”。好家伙，最低音量（6%）我都觉得震耳欲聋，你这样是想让用户报工伤吗？
+
 ## 司法人工智能
 
 ### Large Legal Fictions: Profiling LegalHallucinations in Large Language Models

Solutions/Other-English-LearningNotes-SomeWords.md

@@ -875,7 +875,7 @@ tags: [其他, 知识, 英语, Notes]
 |butchery|n. 残杀，杀戮，屠宰工作|
 |||
 |secretariat|n. 秘书处，书记处|
-|sheriff|n. 县(郡)治安官|
+|<font color="#28bea0" title="二次复习">sheriff</font>|n. 县(郡)治安官|
 |inasmuch|adv. 由于，〈古〉只要<details><summary>例句</summary>Inasmuch as you are an adult, you must be responsible for your own behavior.<br/>因为你是一个成年人，你必须对自己的行为负责。</details>|
 |countenance|n. 面容，脸色，面部表情<br/>v. 支持，赞成，同意|
 |||
@@ -927,6 +927,7 @@ tags: [其他, 知识, 英语, Notes]
 |||
 |snore|n. 鼾声<br/>v. 打鼾|
 |lark|n. 云雀，百灵鸟，玩乐，嬉戏<br/>v. 捉云雀，嬉戏，骑马越野，愚弄|
+|||
 
 <p class="wordCounts">单词收录总数</p>